Why does an analytic function $z^2$, which satisfies the Cauchy-Riemann equations, not satisfy the condition of being harmonic?
We have $u = x^2-y^2$ and $v = 2xy$
\begin{align} du/dx &= 2x, &du/dy &= -2y\\ dv/dx &= 2y, &dy/dy &= 2x \end{align} Which satisfies the condition such that $du/dx = dv/dy$ and $dv/dy = -dv/dx$.
I am asked to prove that $u$ and $v$ are each harmonic functions of $x$ and $y$.
but when I take the second derivative $d^2u/dx^2 = 2$, $d^2u/dy^2 = -2$, which both when added together equal to zero. So $u$ is harmonic.
BUT v is not because $dv^2/dy^2 = 2$ and $dv^2/dx^2$ is 2, which when added together are not equal to 0.
An analytic function should satisfy these conditions and I seem to do it right.
Any hints?
I am asked to prove that u and v are each harmonic functions of x and y.